GRE 9768 #64
From here
How do you prove that? I tried this approach based on a question and solution in the Princeton GRE exam
The question:
The solution:
(Btw, it's supposed to be $f'(y)$ not $f'(x)$?)
Approach:
$$0 \le \lim_{x \to y} \frac{|f(x) - f(y)|}{|x-y|} \le \lim_{x \to y} \frac{E|x-y|}{|x-y|} = E$$
Thus $|f'(y)| = \lim_{x \to y} \frac{|f(x) - f(y)|}{|x-y|}$ exists, namely, it is some constant between $0$ and $E$ and hence $f'(y)$ exists?
If the above can work, why only almost everywhere? Where is compactness used?
Otherwise, how can I approach this please?
Your approach doesn't work, all that you get from the inequalities you wrote is that the difference quotients are bounded, that doesn't imply that the limit exists.
The a.e. differentiability of Lipschitz continuous functions - defined on (open, or otherwise nice enough to talk about differentiability) subsets of $\mathbb{R}^n$ - is Rademacher's theorem, which is not trivial to prove.
For $n = 1$, we have a less mighty hammer available. Lipschitz continuous functions defined on an interval $I \subset \mathbb{R}$ are absolutely continuous, and absolutely continuous functions are almost everywhere differentiable (and they are the integral of their a.e. defined derivative). Not that this hammer is trivial either.
Compactness of the domain plays no role whatsoever for either argument.
As has been mentioned in the comments, an easier example to show that III doesn't need to hold in question 64 would be $f(x) = x^{\alpha}$ for some $\alpha \in (0,1)$.